Consider the polar curve $r=\theta$. What is the equation of the tangent line to the curve $r$ at $\theta=\dfrac{\pi}{2}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $y=\dfrac{\pi}{2}x+\dfrac{\pi}{2}$ (Choice B) B $y=x-\dfrac{\pi}{2}$ (Choice C) C $y=-\dfrac{2}{\pi}x+\dfrac{\pi}{2}$ (Choice D) D $y=-\dfrac{\pi}{2}x+\dfrac{\pi}{2}$
Explanation: The slope of the tangent line at a point is equal to $\dfrac{dy}{dx}$ at that point. In the case of polar curves, we can use the relationship: $\dfrac{dy}{dx}=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)}$ Then we can use the point-slope form to complete the equation for the tangent line through $\theta=\dfrac{\pi}{2}$. For a polar curve, $x={r}\cos(\theta)$ and $y={r}\sin(\theta)$. Therefore, in our problem we have: $\begin{aligned} x&={\theta}\cos(\theta) \\\\ y&={\theta} \sin(\theta) \end{aligned}$ Let's find $\dfrac{dy}{dx}$. $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)} \\\\ &=\dfrac{\sin(\theta)+\theta\cos(\theta)}{\cos(\theta)-\theta\sin(\theta)} \end{aligned}$ Evaluating $\dfrac{dy}{dx}$ at ${\theta = \dfrac{\pi}{2}}$ gives us the slope of our tangent line. $\begin{aligned} {\left. \dfrac{dy}{dx}\right| _{\theta=\tfrac{\pi}{6}}}&=\dfrac{\sin\left({\dfrac{\pi}{2}}\right)+\left({\dfrac{\pi}{2}}\right)\cos\left({\dfrac{\pi}{2}}\right)}{\cos\left({\dfrac{\pi}{2}}\right)-\left({\dfrac{\pi}{2}}\right)\sin\left({\dfrac{\pi}{2}}\right)} \\\\ &=\dfrac{1+\left(\dfrac{\pi}{2}\right)(0)}{0-\dfrac{\pi}{2}(1)} \\\\ &={-\dfrac{2}{\pi}} \end{aligned}$ We now find $x$ and $y$ at the point $\theta=\dfrac{\pi}{2}$. $\begin{aligned} {x\left({\dfrac{\pi}{2}}\right)}&=\left({\dfrac{\pi}{2}}\right)\cos\left({\dfrac{\pi}{2}}\right) \\\\ &=\dfrac{\pi}{2}(0) \\\\ &={0} \\\\ \\\\ y\left({\dfrac{\pi}{2}}\right)}&=\left({\dfrac{\pi}{2}}\right) \sin\left({\dfrac{\pi}{2}}\right) \\\\ &=\dfrac{\pi}{2}(1) \\\\ &=\dfrac{\pi}{2}} \end{aligned}$ Therefore the equation of our tangent line is: $\begin{aligned} y-\dfrac{\pi}{2}}&={-\dfrac{2}{\pi}}(x-{0}) \\\\ y&=-\dfrac{2}{\pi}x+\dfrac{\pi}{2} \end{aligned}$ The graph of the tangent is shown.